Question

Figure shows graph of stopping potential $V_0$ vs. frequency $f$ for two metal surfaces A and B respectively, then

• A. ${\theta }_B>{\theta }_A$
• B. ${\varphi }_B>{\varphi }_A$ [$\varphi$ is magnitude of work function of metal]
• C. ${\theta }_A={\theta }_B$
• D. ${\varphi }_A>{\varphi }_B$

Answer 1

Answer: (B), (C)

The correct options are
B ${\varphi }_B>{\varphi }_A$ [$\varphi$ is magnitude of work function of metal]
C ${\theta }_A={\theta }_B$

Stopping voltage $V_0$ is $\Rightarrow e{V}_0=hf-\varphi$

$V_0=\frac{hf}{e}-\frac{\varphi }{e}$, where $\frac{h}{e}$ is slope which is constant, so ${\theta }_A={\theta }_B$

Also, from the graph, the magnitude of y intercept of B is more than A. So,
${\varphi }_B>{\varphi }_A$.